Question

An air-filled balloon has a volume of 225 L at 0.940 atm and 25 °C. Soon after, the pressure changes to 0.990 atm and the temperature changes to 0 °C. What is the new volume of the balloon?

Answer 1

The right choice is V₂ = 195.7 L

Step-by-step explanation:

• We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V: is the volume of the gas in L.

n: is the no. of moles of the gas in mol.

R: is the general gas constant,

T: is the temperature of the gas in K.

If n is constant, and have different values of P, V and T:

(P₁*V₁) / T₁ = (P₂ * V₂) / T₂

Knowing that:

V₁ = 225 L , P₁ = 0.940 atm

T₁ = 25 °C + 273 = 298 K

V₂ = ??? L, P₂ = 0.990 atm

T₂ = 0 °C + 273 = 273 K

applying in the above equation

(P ₁* V₁) / T₁ = (P₂ * V₂) / T₂

(0.940 atm * 225 L) / 298 K= (0.990 atm * V₂) / 273 K

V₂ = (0.940 atm * 225 L * 273 K) / (298 K * 0.990 atm)

V₂ = 195.7 L

So, the right choice is:

V₂ = 195.7 L